Problem: The grades on a history midterm at Santa Rita are normally distributed with $\mu = 79$ and $\sigma = 3.5$. Umaima earned a $69$ on the exam. Find the z-score for Umaima's exam grade. Round to two decimal places.
A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Umaima's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{69 - {79}}{{3.5}}} $ ${ z \approx -2.86}$ The z-score is $-2.86$. In other words, Umaima's score was $2.86$ standard deviations below the mean.